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Question Number 27604 by 0722136841 last updated on 10/Jan/18

$$\mathrm{Let}$$$$S_n=\frac{1}{1^3}+\frac{1+2}{1^3+2^3}+...+\frac{1+2+...+n}{1^3+2^3+...+n^3};n=1,2,3,..$$$$\mathrm{Then}{S}_n\mathrm{is}\mathrm{greater}\mathrm{than}$$

Commented by abdo imad last updated on 11/Jan/18

$$S_n=\sum _{k=1}^n\frac{\sum _{p=1}^{k}p}{\sum _{p=1}^k{p}^3}letsimplify{S}_n$$$$weknowthat\sum _{p=1}^{k}p=\frac{k(k+1)}{2}and\sum _{p=1}^k{p}^3={\left(\frac{k(k+1)}{2}\right)}^2$$$$S_n=\sum _{k=1}^n\frac{2}{k(k+1)}=2\sum _{k=1}^n(\frac{1}{k}-\frac{1}{k+1})$$$$S_n=2(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n}-\frac{1}{n+1})=2(1-\frac{1}{n+1})$$$$=\frac{2n}{n+1}butn+1<2nforalln>1\Rightarrow \frac{1}{n+1}>\frac{1}{2n}$$$$\Rightarrow \frac{2n}{n+1}>\frac{1}{2}\Rightarrow S_n>\frac{1}{2}andalsowehave\frac{1}{2}<S_n<2$$$$and{lim}_{n->\propto }{S}_n=2$$

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